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Proof step 2 from proof ocs-del1 (previous step)

1 < #oc(a, x) ⊦ a ∈ x -1l a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ x ⊦ #oc(a, x) = 1, a ∈ x -1l a

Rule argument

Additional information

Used simplifier rules: oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%in-05: ⊦ ¬ a ∈ y -1l b ∧ a ∈ y → a = b; (forward) from specification list-del% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)